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EXPLANATION OF TIN IONIZATIONS
By Prof. L. Kaliambos (Natural Philosopher in New Energy) June 8 , 2015 Tin is a chemical element with symbol Sn and atomic number 50. However despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of many electon atoms related to the chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. (EXPERIMENTS REJECT RELATIVITY). It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light, which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin. So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this correct image of Tin including the following ground state electron configuration: 1s22s22p63s23p63d104s24p64d105s25px15py1 According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies (eV) of tin (from (E1 to E5 ) are the following: E1 = 7.34 , E2 = 14.63, E3 = 30.5, E4 = 40.735 and E5 = 72.28 . For understanding better such ionization energies see also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS. Moreover in “User Kaliambos” you can see my paper “ Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” published in Ind. J. Th. Phys. (2008). EXPLANATION OF E1 = 7.34 eV = -E(5px1) Here the E(5px1) represents the binding energy of the first outer electron, the 5px1. The charges (-48e) of (1s22s22p63s23p63d104s2 4p64d10 5s2) screen the nuclear charge (+50e) and for a perfect screening we would have ζ = 2. However the two electrons of 5p2 repel the 5s2 and 4d10 and lead to the deformations of electron clouds. Thus ζ > 2. Since 5px1 is the first outer electron of parallel spin we apply the Bohr formula to write E1 = -E(5px1) = 7.34 eV = - (-13.6057)ζ2/n2 Therefore E1 = 7.34 eV = (13.6057)ζ2 / 52 Then we get ζ = 3.67 > 2 . ' ' EXPLANATION OF E2 = 14.63 eV = - E(5py1) Here the E(5py1) represents the binding energy of the second outer electron, the 5py1. As in the case of E1 the charges (-48e) of (1s22s22p63s23p63d104s2 4p64d10 5s2) screen the nuclear charge (+50e) and for a perfect screening we would have the same ζ = 2. However the electron of 5py1 repels the 5s2 and the 4d10 and leads to the deformations of electron clouds. Thus ζ > 2. Since 5py1 is the second outer electron of parallel spin we apply the Bohr formula to write E2 = -E(5py1) = 14.63 eV = - (-13.6057)ζ2/n2 Therefore E2 = 14.63 eV = (13.6057)ζ2 / 52 Then we get ζ = 5.19 > 2 . Here 5.19 > 3.67 > 2 means that the second outer electron breaks more the symmetry and leads to a great deformation of electron clouds. Note that in the case of E1 both 5px1 and 5py1 exert mutual electric and magnetic repulsions. Under this condition in the case of E1 they repel from symmetrical positions the 5s2 and 4d10 and lead to a less deformation of electron clouds. EXPLANATION OF E3 = 30.5 eV = -E(5s2) + E(5s1) Here the E(5s2) represents the binding energy of 5s2, while the E(5s1) represents the binding energy of 5s1. The charges (-46e) of (1s22s22p63s23p63d104s24p64d10 ) screen the nuclear charge (+50e) and for a perfect screening we would have ζ = 4. However the electrons of 5s penetrate the 4d10 and lead to the deformations of electron clouds. Thus ζ > 4. Note that 5s2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(5s2) = -+ (16.95)ζ - 4.1 / n2 On the other hand, since 5s1 consists of one electron, we apply the Bohr formula to write E(5s1) = (-13.6057)ζ2/n2 Therefore E3 = 30.5 eV = -E(5s2) + E(5s1) = - (16.95)ζ + 4.1) / n2 Since n = 5 the above equation can be written as (13.6057)ζ2 - (16.95) ζ - 758.4 = 0 Then solving for ζ we get ζ = 8.12 > 4 . Of course the two electrons of opposite spin (5s2) do not provide any mutual repulsion, because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion giving a vibration energy. This situation indeed occurs, because the peripheral velocity of a spinning electron is faster than the speed of light, which invalidates Einstein’s theory of special relativity. (See my FASTER THAN LIGHT). However under the influence of invalid relativity and in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of opposite spin today many physicists believe incorrectly that it is due to the Coulomb repulsion. Under such fallacious ideas I published my paper of 2008. ' ' EXPLANATION OF E4 = 40.735 eV = -E(5s1) ' Here the E(5s1) represents the binding energy of 5s1. As in the case of E3 the charges (-46e) of (1s22s22p63s23p63d104s2 4p64d10 ) screen the nuclear charge (+50e) and for a perfect screening we would have the same ζ = 4. However the electron of 5s1 penetrates the 4d10 and leads to the deformations of electron clouds. Thus ζ > 4. Since 5s1 consists of one electron, we apply the Bohr formula to write E4 = - E(5s1) = 40.735 eV = - (-13.6057)ζ2/n2 Therefore E4 = 40.735 eV = (13.6057)ζ2 / 52 Then we get ζ = 8.65 > 4 . Here the 8.65 > 8.12 > 4 means that the one electron (5s1) breaks more the symmetry of spherical shells than the two electrons (5s2) and leads to greater deformations of spherical electron clouds. ' ''' '''EXPLANATION OF Ε5 = 72.28 eV = -E(4d10) + E(4d9 ) Here the E(4d10) represents the binding energy of the 10 electrons (4d10), while the E(4d9) represents the binding energy of 9 electrons (4d9) .The charges (-36e) of (1s22s22p63s23p63d104s24p6) screen the nuclear charge (+50e) and for a perfect screening we would have an effective ζ = 14. However, in this case, the experiments of ionizations show that each pair of 4d repels not only the electrons of 4p6 but also the electrons of 4d. Surprisingly such repulsions occur in such a way that ζ < 14. Note that 4d10 consists of five pairs (10 electrons with opposite spin) Thus for the five pairs of opposite spin we apply my formula of 2008. Under this condition one may write -E(4d10) = -5+ (16.95)ζ - 4.1 / n2 On the other hand the 4d9 consists of four pairs (8 electrons with opposite spin ) whose the binding energy is given by applying my formula of 2008, and of one electron whose the binding energy is given by applying the Bohr formula. So we may write E(4d9 ) = 4+ (16.95)ζ - 4.1 / n2 + (13.6057)ζ2 / n2 Therefore Ε5 = 72.28 eV = -E(4d10) + E(4d9 ) = - (16.95)ζ + 4.1 / n2 It is of interest to note that in this sub- shell of 4d the experiments of ionizations showed that 4 < n < 5. (See my paper EXPLANATION OF TECHNETIUM IONIZATIONS). Thus using n = 4 one writes (13.6057)ζ2 - (16.95)ζ - 1152.38 = 0 and solving for ζ we get ζ = 9.85 < 14 Whereas using n = 5 one writes (13.6057)ζ2 - (16.95)ζ - 1802.9 = 0 and solving for ζ we get ζ = 12.15 < 14 . Category:Fundamental physics concepts